Determine the most general Mobius transform that satisfies:
i) two fixed points z=0 and z=1;
ii) maps the unit circle $|z|=1$ and the bisector of first and third quarter to lines.
My guess is: $$\frac{cz(1 \pm \frac{1}{\sqrt{2}}(1+i))}{cz \pm \frac{c}{\sqrt{2}}(1+i)}$$ where $c \in \mathbb{C}$
May you tell me if it's correct?
edit: In the (ii) i want to say that the Mobius $$\frac{az+b}{cz+d}$$ transformations takes the points of the complex plane $|z|=1$ and $z(t) = t e^{i \frac{\pi}{4}}$ in lines (so I infer that the point $-\frac{d}{c}$ is one of the two points of intersection between $|z|=1$ and $z(t) = t$.
In order to satisfy condition (i) we can write the transformation as $$w=\frac{az}{z+a-1}$$ for some $a\in\mathbb{C}$
We can rearrange and write this as $$z=\frac{w-aw}{w-a}$$
In order to apply the conditions in (ii), we can set $w=u+iv$ and $a=p+iq$
Firstly, the locus $|z|=1$ becomes $|w-aw|=|w-a|$ which is equivalent to $$(u(1-p)+qv)^2+(v(1-p)-qu)^2=(u-p)^2+(v-q)^2$$
For this to represent a line, we require the coefficients of $u^2$ and $v^2$ to be zero, each of which give the relation $$(1-p)^2+q^2=1$$
In addition the coefficient of $uv$ is identically zero.
Secondly, we require the image of the locus $Re(z)=Im(z)$ also to be a line, and after a few lines of algebra, we get $$q=p-1$$
So the possible values of $a$ are $$1\pm\frac{1}{\sqrt{2}}(1+i)$$